The equation of the normal to the curve $y=x \log x$ parallel to $2 x-2 y+3=0$ is

The equation of the normal to the curve $y=x \log x$ parallel to $2 x-2 y+3=0$ is
  1. $x+y=3 \mathrm{e}^{-2}$
  2. $x-y=3 \mathrm{e}^{-2}$
  3. $x-y=3 \mathrm{e}^2$
  4. $x+y=3 \mathrm{e}^2$

Solution

$\begin{aligned} & y=x \log x ...(i)\\ \therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=1+\log x \end{aligned}$
Slope of the normal $=\frac{-1}{\frac{\mathrm{~d} y}{\mathrm{~d} x}}=-\frac{1}{1+\log x}$ Slope of the given line is 1. Since the normal is parallel to the given line. $\begin{aligned} \therefore \quad & -\frac{1}{1+\log x}=1 \\ & \Rightarrow \log x=-2 \\ & \Rightarrow x=\mathrm{e}^{-2} \\ & y=-2 \mathrm{e}^{-2} \end{aligned}$ ...[From (i)] $\therefore \quad$ Equation of the normal at $\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)$ is $\begin{aligned} & y+2 \mathrm{e}^{-2}=1\left(x-\mathrm{e}^{-2}\right) \\ & \Rightarrow y+2 \mathrm{e}^{-2}=x-\mathrm{e}^{-2} \\ & \Rightarrow x-y=3 \mathrm{e}^{-2} \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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